INTRODUCTION TO ELASTICITY David Roylance Department of Materials Science and Engineering Massachusetts Institute of Technology Cambridge, MA 02139 January 21, 2000 Introduction This module outlines the basic mechanics of elastic response — a physical phenomenon that materials often (but do not always) exhibit. An elastic material is one that deforms immediately upon loading, maintains a constant deformation as long as the load is held constant, and returns immediately to its original undeformed shape when the load is removed. This module will also introduce two essential concepts in Mechanics of Materials: stress and strain. Tensile strength and tensile stress Perhaps the most natural test of a material’s mechanical properties is the tension test, in which a strip or cylinder of the material, having length L and cross-sectional area A, is anchored at one end and subjected to an axial load P – a load acting along the specimen’s long axis – at the other. (See Fig. 1). As the load is increased gradually, the axial deflection δ of the loaded end will increase also. Eventually the test specimen breaks or does something else catastrophic, often fracturing suddenly into two or more pieces. (Materials can fail mechanically in many different ways; for instance, recall how blackboard chalk, a piece of fresh wood, and Silly Putty break.) As engineers, we naturally want to understand such matters as how δ is related to P , and what ultimate fracture load we might expect in a specimen of different size than the original one. As materials technologists, we wish to understand how these relationships are influenced by the constitution and microstructure of the material. Figure 1: The tension test. One of the pivotal historical developments in our understanding of material mechanical properties was the realization that the strength of a uniaxially loaded specimen is related to the 1 magnitude of its cross-sectional area. This notion is reasonable when one considers the strength to arise from the number of chemical bonds connecting one cross section with the one adjacent to it as depicted in Fig. 2, where each bond is visualized as a spring with a certain stiffness and strength. Obviously, the number of such bonds will increase proportionally with the section’s area1. The axial strength of a piece of blackboard chalk will therefore increase as the square of its diameter. In contrast, increasing the length of the chalk will not make it stronger (in fact it will likely become weaker, since the longer specimen will be statistically more likely to contain a strength-reducing flaw.) Figure 2: Interplanar bonds (surface density approximately 1019 m−2). Galileo (1564–1642)2 is said to have used this observation to note that giants, should they exist, would be very fragile creatures. Their strength would be greater than ours, since the cross-sectional areas of their skeletal and muscular systems would be larger by a factor related to the square of their height (denoted L in the famous DaVinci sketch shown in Fig. 3). But their weight, and thus the loads they must sustain, would increase as their volume, that is by the cube of their height. A simple fall would probably do them great damage. Conversely, the “proportionate” strength of the famous arachnid mentioned weekly in the SpiderMan comic strip is mostly just this same size effect. There’s nothing magical about the muscular strength of insects, but the ratio of L2 to L3 works in their favor when strength per body weight is reckoned. This cautions us that simple scaling of a previously proven design is not a safe design procedure. A jumbo jet is not just a small plane scaled up; if this were done the load-bearing components would be too small in cross-sectional area to support the much greater loads they would be called upon to resist. When reporting the strength of materials loaded in tension, it is customary to account for this effect of area by dividing the breaking load by the cross-sectional area: Pf σf = (1) A0 where σf is the ultimate tensile stress, often abbreviated as UTS, Pf is the load at fracture, and A0 is the original cross-sectional area. (Some materials exhibit substantial reductions in cross-sectional area as they are stretched, and using the original rather than final area gives the so-call engineering strength.) The units of stress are obviously load per unit area, N/m2 (also 1 The surface density of bonds NS can be computed from the material's density ρ, atomic weight Wa and 2=3 Avogadro's number NA as NS =(ρNA=Wa) . Illustrating for the case of iron (Fe): 2 g 23 atoms 3 7:86 cm3 · 6:023 × 10 mol 15 atoms NS = =1:9×10 55:85 g cm2 mol 15 atom NS ≈ 10 cm2 is true for many materials. 2Galileo, Two New Sciences, English translation by H. Crew and A. de Salvio, The Macmillan Co., New York, 1933. Also see S.P. Timoshenko, History of Strength of Materials, McGraw-Hill, New York, 1953. 2 Figure 3: Strength scales with L2, but weight scales with L3. called Pascals, or Pa) in the SI system and lb/in2 (or psi) in units still used commonly in the United States. Example 1 In many design problems, the loads to be applied to the structure are known at the outset, and we wish to compute how much material will be needed to support them. As a very simple case, let’s say we wish to use a steel rod, circular in cross-sectional shape as shown in Fig. 4, to support a load of 10,000 lb. What should the rod diameter be? Figure 4: Steel rod supporting a 10,000 lb weight. Directly from Eqn. 1, the area A0 that will be just on the verge of fracture at a given load Pf is Pf A0 = σf All we need do is look up the value of σf for the material, and substitute it along with the value of 10,000 lb for Pf , and the problem is solved. A number of materials properties are listed in the Materials Properties module, where we find the UTS of carbon steel to be 1200 MPa. We also note that these properties vary widely for given materials depending on their composition and processing, so the 1200 MPa value is only a preliminary design estimate. In light of that uncertainty, and many other potential ones, it is common to include a “factor of safety” in the design. Selection of an appropriate factor is an often-difficult choice, especially in cases where weight or cost restrictions place a great penalty on using excess material. But in this case steel is 3 relatively inexpensive and we don’t have any special weight limitations, so we’ll use a conservative 50% safety factor and assume the ultimate tensile strength is 1200/2 = 600 Mpa. We now have only to adjust the units before solving for area. Engineers must be very comfortable with units conversions, especially given the mix of SI and older traditional units used today. Eventually, we’ll likely be ordering steel rod using inches rather than meters, so we’ll convert the MPa to psi rather than convert the pounds to Newtons. Also using A = πd2/4 to compute the diameter rather than the area, we have 1 2 4A 4P 4 × 10000(lb) d = = f = =0.38 in 2 π πσf 2 6 2 −4 lb=in 3 r s π × 600 × 10 (N/m ) × 1.449 × 10 N=m2 4 5 We probably wouldn’t order rod of exactly 0.38 in, as that would be an oddball size and thus too expensive. But 3/800 (0.375 in) would likely be a standard size, and would be acceptable in light of our conservative safety factor. If the specimen is loaded by an axial force P less than the breaking load Pf ,thetensile stress is defined by analogy with Eqn. 1 as P σ = (2) A0 The tensile stress, the force per unit area acting on a plane transverse to the applied load, is a fundamental measure of the internal forces within the material. Much of Mechanics of Materials is concerned with elaborating this concept to include higher orders of dimensionality, working out methods of determining the stress for various geometries and loading conditions, and predicting what the material’s response to the stress will be. Example 2 Figure 5: Circular rod suspended from the top and bearing its own weight. Many engineering applications, notably aerospace vehicles, require materials that are both strong and lightweight. One measure of this combination of properties is provided by computing how long a rod of the material can be that when suspended from its top will break under its own weight (see Fig. 5). Here the stress is not uniform along the rod: the material at the very top bears the weight of the entire rod, but that at the bottom carries no load at all. To compute the stress as a function of position, let y denote the distance from the bottom of the rod and let the weight density of the material, for instance in N/m3, be denoted by γ. (The weight density is related to the mass density ρ [kg/m3]byγ=ρg,whereg=9.8m/s2 is the acceleration due to gravity.) The weight supported by the cross-section at y is just the weight density γ times the volume of material V below y: W(y)=γV = γAy 4 The tensile stress is then given as a function of y by Eqn.